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I am having problems understanding how this equation works in c:

char *sum(char *a, int b) {
    return &a[b];

printf("%d", sum(5, 4));

I understand how arrays work, and I understand how to reference and de-reference a variable to a memory location, but I don't understand where the addition comes in to play here.

It makes sense to me that return &a[4] for example would just return a de-referenced non-existent memory location and cause an error.

Can someone explain this to me in easy terminology?

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up vote 3 down vote accepted

In C, the array index operation a[b] is implicitly treated by the compiler as *(a + b).

This property means that you can write, for example:

assert(5["A string"] == 'i');

And it is true.

Because C also performs implicit conversions from int to a pointer type, your example is evaluated like this:

sum(5,4) -> char *a = 5, b = 4
return &a[b]; -> return &(*(a + b)); -> return &(*((char*)5 + 4));

Because there is no memory access, (the value of a[b] is not inspected or assigned) the & and * operators are canceled out, and the inner sum (5+4) is cast to char* returned as-is.

The particular location of a in memory is irrelevant, and the value in a is irrelevant because no memory access is performed.

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Thank you, that makes sense. I guess I had never seen that before in any other language. – Jonathan Nov 28 '13 at 5:54
@Jonathan: You won't see this in other languages except perhaps C++ (where the rules are a bit different). This is a peculiarity unique to C. – greyfade Nov 28 '13 at 17:29

It will return the address of b char units in memory from the pointer to a.

As it will return an address to memory, you can find whatever in that address, a NULL or any other character.

So if you have sum(5,4), the first parameter will be taken as an address (the starting point) and it will return 4 char units in memory from 5.

  • If the machine defines a char as 1 byte, then it will return 9.
  • If the machine defines a char as 2 bytes, then it will return 5 + 2 + 2 + 2 + 2 = 13.
  • And so on.
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'4 char units in memory'? If the 'char * a' is 'bff57400', wouldn't 4 'char units' be '7'? – Jonathan Nov 28 '13 at 0:58
if bff57400 is an address, &a[4] is bff57404 (if a char is 1 byte),. If its an array of char 'b','f','f'.... then a[4] is 7. – mattnz Nov 28 '13 at 2:02

You are nearly there. &a[4] returns a + 4 times sizeof(type of elements in a) (in this case char). This is then returned, in you case as 5+4 = 9.

The value is valid, but the deference to invalid memory only happens if you use it to access memory, so at this point nothing "bad" has happened.

char *a = sum(5,4) 
printf ("%d",a);       is OK and prints 9 
printf ("%d", *a);     all bets off..... 

another example to think about : char *x = NULL; x is NULL, but you do not get a Null pointer exception thrown till you do something with it such as *x = y ;

I also suggest you compile this code with -Wall and study the warnings.

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I'm still confused. Let's say the address of 'a' is 'bff57400'. That would mean a[4] would return '7', which referenced will have a completely different memory location. Where does the addition come in? – Jonathan Nov 28 '13 at 0:55
The variable a, which happens to be stored as an address, has the value 5 based on how sum is invoked. bff57400 is a red herring. – James McLeod Nov 28 '13 at 4:05

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